University Quantum Field Theory I · The One-Loop Vertex: the Anomalous Magnetic Moment and Infrared Divergences · 14.7

Soft bremsstrahlung: the eikonal emission probability

Every scattered charge radiates, and the soft end of that radiation takes the same form for an electron, a pion or a classical charge. Here you derive the factor that fixes it, turn it into a photon count that grows without limit as the photon mass goes to zero, and hand that logarithm to topic 14.08's measured rate.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Attach a photon of momentum k to an external line and the propagator beside it carries 1/(2p⋅k), which grows without bound as k → 0 while everything else in the diagram stays finite. In that soft limit the Dirac equation turns the numerator into 2p^μ, and the amplitude factorises into the elastic one times e[p′⋅ε*/(p′⋅k) − p⋅ε*/(p⋅k)]. Spin, the target and the mechanism of the scattering have all dropped out: the bracket is the Fourier transform of the current of a classical charge whose four-velocity jumps from p/m to p′/m, which is why Maxwell's equations already knew the answer.

The bracket is a conserved current, so the polarisation sum may use −g_μν; the result falls as 1/ω² and crowds into cones along the incoming and outgoing lines. Times ω², its average over directions is exactly 2f_IR(q²), the function that multiplied the photon-mass logarithm in F₁, and the phase space d³k/(2ω) leaves dN/dω = (2α/π)f_IR/ω: the same number of photons in every e-fold of energy. Between a photon mass μ_γ and a resolution Eₗ that is (α/π)f_IR ln(Eₗ²/μ_γ²), divergent as μ_γ → 0 while the energy carried stays finite.

The model stops when the photon stops being soft: as ω approaches |q| or E the hard amplitude feels k, and internal lines and the magnetic moment radiate; and once the logarithm drives the order-α number towards 1, one photon is no longer the right question.

Simple definition: Soft bremsstrahlung is the emission, from the external lines of a scattering, of photons far softer than the momentum transfer; its amplitude is the elastic amplitude times an eikonal factor fixed only by the charges and momenta of those lines.
Example: For 1 GeV electrons scattered through 30° by a heavy nucleus, f_IR = 12.84, so each e-fold of photon energy holds 0.060 photons: 0.137 photons between 1 and 10 MeV per scatter, carrying 0.54 MeV between them.

The eikonal factorM(p → p′ + γ) ≈ M₀(p → p′) × e[p′⋅ε*/(p′⋅k) − p⋅ε*/(p⋅k)]

Derived from the 1/(2p⋅k) poles of the two external legs. Spin, the target and whatever caused the scattering drop out; only charge and four-momentum remain.

k, ε: photon momentum in GeV and polarisation; the factor is in GeV⁻¹; ħ = c = 1, metric (+, −, −, −); for ω ≪ |q|, E

Polarisation sum for a conserved currentΣ|ε*⋅J|² = −J⋅J = 2p⋅p′/((p⋅k)(p′⋅k)) − m²/(p′⋅k)² − m²/(p⋅k)²

Derived. Topic 8.07's −g_μν sum is legal only because the two legs together conserve charge; one leg alone would give −m²/(p⋅k)², a negative rate.

J^μ = p′^μ/(p′⋅k) − p^μ/(p⋅k) in GeV⁻¹, with k⋅J = 1 − 1 = 0; m the electron mass in GeV

The angular integral is f_IR∫ dΩ/(4π) ω²(−J⋅J) = 2f_IR(q²)

Derived with one Feynman parameter; each leg's own term contributes −1. It is topic 14.06's function, here produced by real photons rather than virtual ones.

f_IR = ∫₀¹ dξ (m² − q²/2)/(m² − ξ(1 − ξ)q²) − 1, dimensionless; q = p′ − p in GeV; Ω the photon's direction

f_IR in closed formf_IR(q²) = χ coth χ − 1, with cosh χ = p⋅p′/m² = 1 − q²/(2m²)

Derived from the ξ integral. Small χ gives −q²/(3m²), large χ gives ln(−q²/m²) − 1; 1 GeV electrons at 30° have χ = 13.84 and f_IR = 12.84.

χ the rapidity of p′ seen from the rest frame of p, dimensionless; m and q in GeV

The soft-photon spectrumdN/dω = (2α/π) f_IR(q²)/ω, for μ_γ ≪ ω ≪ |q|, E

Derived. Equal photon numbers in every e-fold of energy: 0.060 per e-fold for 1 GeV electrons at 30°, the flat spectrum classical radiation also gives.

N the mean photon number, ω the photon energy in MeV, α = 1/137.036; the energy spectrum ω dN/dω is flat

Emission probability below the resolutionP(μ_γ ≤ ω ≤ Eₗ) = (α/π) f_IR(q²) ln(Eₗ²/μ_γ²)

Derived. It diverges as μ_γ → 0 with the coefficient of topic 14.06's virtual logarithm, opposite in sign, while the energy carried, (2α/π)f_IR Eₗ, stays finite.

μ_γ the photon-mass regulator and Eₗ the detector resolution, both in MeV; leading logarithm, order α

The emitting propagator sits beside its mass shell

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). An electron scatters from p to p′, q = p′ − p, and a photon of momentum k, energy ω and polarisation ε leaves the outgoing line. The propagator between the hard vertex and the emission carries p′ + k, and (p′ + k)² − m² = 2p′⋅k + μ_γ², with μ_γ topic 8.08's photon mass. For 1 GeV electrons scattered through 30°, a 1 MeV photon at 90° to p′ gives 2p′⋅k = 2 × 10⁻³ GeV², 134 times below |q²| = 0.268 GeV², and along p′ only 2.6 × 10⁻¹⁰ GeV². In the numerator ū(p′)γ^ν(p̸′ + m) = ū(p′)[2p′^ν − (p̸′ − m)γ^ν] = 2p′^ν ū(p′) by the Dirac equation, the k̸ dropped being one power of ω down. So the outgoing leg gives e p′⋅ε*/(p′⋅k), and the incoming leg, whose propagator carries p − k and the denominator −2p⋅k, gives −e p⋅ε*/(p⋅k), both derived. Photons from inside the hard process, and the k-dependence of M₀ itself, enter only at relative order ω/|q|.

Only charge and momentum survive: a classical current

The reduction used nothing but the Dirac equation, and a charged scalar gives the same factor at once: its vertex −ie(2p′ + k)^ν becomes −2ie p′^ν beside the same pole i/(2p′⋅k). What spin adds is the magnetic coupling, through σ^νλ k_λ, which carries a power of k and so enters at order ω⁰, one power of ω below the eikonal 1/ω. Nor does the factor know what scattered the electron: M₀ is evaluated at k = 0 and simply multiplies through, whether the kick came from a nucleus, another electron or a weak boson. The factor is classical. A point charge with four-velocity p/m before t = 0 and p′/m after carries the current j^μ(x) = e∫dτ u^μ δ⁴(x − x(τ)), and with a convergence factor at large |τ| its Fourier transform is j̃^μ(k) = ie[p′^μ/(p′⋅k) − p^μ/(p⋅k)], derived in two lines of integration. The soft amplitude is the amplitude for the radiation field of that classical kick, the picture topic 14.09 turns into a coherent state.

A conserved current, two cones and a notch on each leg

Write the bracket as ε*⋅J with J^μ = p′^μ/(p′⋅k) − p^μ/(p⋅k), in GeV⁻¹. Then k⋅J = 1 − 1 = 0, derived: each leg alone gives exactly 1, and only their difference vanishes, the statement that the charge leaving equals the charge that arrived. That is what topic 8.07 needs to replace Σ ε_μ ε*_ν by −g_μν, giving −J⋅J = 2p⋅p′/((p⋅k)(p′⋅k)) − m²/(p′⋅k)² − m²/(p⋅k)². One leg alone would give −m²/(p⋅k)², negative: a current that is not conserved cannot be squared with −g_μν. With k = ω(1, n̂), a single leg radiates as v² sin²ψ/(ω²(1 − v cos ψ)²), ψ the angle from the leg: zero exactly along it, about γ²/ω² near ψ = 1/γ, and 4/(ψ²ω²) beyond. For 1 GeV electrons γ = 1957, so each cone is 0.03° wide. For slow charges the legs merge into a dipole, ω²(−J⋅J) → |Δv|² sin²ψ_q, with ψ_q the angle from q.

The angular integral is f_IR, the same in every frame

With k = ω(1, n̂), d³k/(2ω) = (ω/2) dω dΩ and ω²(−J⋅J) depends on n̂ alone, so the angles separate. For timelike P, ∫dΩ/(4π) 1/(P⋅n)² = 1/P², from ½∫dc/(P⁰ − |P|c)² over c from −1 to 1, so each leg's own term gives m²/m² = 1. The cross term takes one Feynman parameter, 1/(ab) = ∫₀¹ dξ/[ξa + (1 − ξ)b]², which turns it into 1/(P_ξ⋅n)² with P_ξ = ξp + (1 − ξ)p′ and P_ξ² = m² − ξ(1 − ξ)q². Hence ∫dΩ/(4π) ω²(−J⋅J) = ∫₀¹ dξ (2m² − q²)/(m² − ξ(1 − ξ)q²) − 2 = 2f_IR(q²), derived: topic 14.06's function, now from real photons. The result depends only on p⋅p′ and m², so it is the same in every frame. The ξ integral is elementary: f_IR = χ coth χ − 1 with cosh χ = p⋅p′/m², χ the rapidity of p′ seen from the rest frame of p. For 1 GeV electrons at 30°, −q²/m² = 1.026 × 10⁶, χ = 13.84 and f_IR = 12.84, within 10⁻⁵ of ln(−q²/m²) − 1.

The energy integral: a logarithm only the photon mass stops

What remains is ∫dω/ω, so dN/dω = (2α/π)f_IR(q²)/ω, derived: equal photon numbers in every e-fold of energy, and a flat energy spectrum ω dN/dω, as classical radiation from a collision of duration τ is flat below 1/τ. With topic 8.08's photon mass, k⁰ = √(|k|² + μ_γ²) and the denominators become 2p′⋅k + μ_γ² and −(2p⋅k − μ_γ²); below |k| ≈ μ_γ the integrand stops growing, so at leading logarithm the mean number between μ_γ and a resolution Eₗ is (α/π)f_IR ln(Eₗ²/μ_γ²), derived. For 1 GeV electrons at 30°, (α/π)f_IR = 0.0298: 0.137 photons between 1 and 10 MeV. With Eₗ = 10 MeV the count reaches 1 at μ_γ = 0.52 eV and grows without limit below, while the energy below 10 MeV stays (2α/π)f_IR Eₗ = 0.60 MeV. The coefficient of ln μ_γ² is −(α/π)f_IR, equal and opposite to its coefficient in topic 14.06's elastic rate; adding the two is topic 14.08's measured rate.

Checks, limits and where the soft factor stops

Dimensions: each leg's factor e p⋅ε*/(p⋅k) is in GeV⁻¹, as an amplitude with one more external particle must be, and d³k/(2ω) supplies GeV², so the probability is a pure number. Slow limit: f_IR → −q²/(3m²) = |Δv|²/3, so ω dN/dω → (2α/(3π))|Δv|², the classical spectrum of a sudden velocity change, quoted; a 10 keV electron turned back by a nucleus has f_IR = 0.0515 against |Δv|²/3 = 0.0507, the 1.6% gap of order v². Fast limit: f_IR → ln(−q²/m²) − 1, a collinear logarithm from the cones that diverges as m → 0. The factor fails once ω nears |q| or E, where M₀ feels k and internal lines and the magnetic moment radiate; Low's theorem, named, gives that next order. For any process the factor is Σᵢ ηᵢeᵢ pᵢ⋅ε*/(pᵢ⋅k), with ηᵢ = +1 for outgoing and −1 for incoming lines, gauge invariant only if charge is conserved: Weinberg's soft-photon theorem, named. Measured: elastic peaks in electron scattering carry radiative tails, and the Mo-Tsai corrections built on this factor are applied to such data, quoted.

Change one variable at a time

Make the relationship visible.

Interactive model

Momentum of the charge p/m6.0

Scattering angle θ60 °

Decades from μ_γ up to Eₗ12

Start at p/m = 0.2: the solid curve lies on the dashed dipole, zero along ±q. Raise p/m and two cones open along p and p′, each notched just outside its leg and peaked near 1/γ, lifting the average 2f_IR. Then add decades and watch the bar pass its frame at large p/m and θ.

f_IR = χ coth χ − 12.642

LARGE −q² FORM ln(−q²/m²) − 12.584

DIPOLE FORM −q²/(3m²)12.000

P(μ_γ TO Eₗ), ORDER α0.339

Live interpretationf_IR = χ coth χ − 1: 2.642. LARGE −q² FORM ln(−q²/m²) − 1: 2.584. DIPOLE FORM −q²/(3m²): 12.000. P(μ_γ TO Eₗ), ORDER α: 0.339

Catch the common trap

Explain before calculating.

Misconception check

It is tempting to read a divergent emission probability as a divergent radiated energy, or as proof that the calculation has gone wrong. It is neither. The spectrum dN/dω = (2α/π)f_IR/ω puts equal numbers of photons in every e-fold of energy, so their number grows as ln(1/μ_γ) while their energy, (2α/π)f_IR Eₗ, is finite and independent of μ_γ: 0.60 MeV below 10 MeV for 1 GeV electrons at 30°, however small the photon mass. Classically the same kick radiates a flat spectrum ω dN/dω, and infinitely many quanta of vanishing energy is what a flat spectrum looks like when counted in photons. What fails is reading (α/π)f_IR ln(Eₗ²/μ_γ²) as the probability of exactly one photon. It is a mean number, and once it nears 1, at μ_γ = 0.52 eV in that example, order-α perturbation theory has stopped describing exclusive final states; topics 14.08 and 14.09 say what can be measured.

In soft bremsstrahlung from an electron kicked from p to p′, the photon polarisation sum is done with −g_μν, although only two polarisations are physical. Why is that allowed?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyElectrons of 2.0 GeV scatter through 20° from a heavy nucleus that does not recoil. Find f_IR, the mean number of photons per scatter with energies between 5 and 50 MeV, and the energy those photons carry.

Recoil-free scattering keeps E′ = E, so −q² = 4|p|² sin²(θ/2) = 4 × (2.0 GeV)² × sin²10° = 0.4825 GeV² and |q| = 695 MeV. Photons up to 50 MeV are far below |q| and E, so the soft factor applies.
−q²/m² = 0.4825/(0.000511)² = 1.848 × 10⁶, far above 1, so f_IR = ln(−q²/m²) − 1 = 14.43 − 1 = 13.43; the closed form χ coth χ − 1 agrees to the digits shown.
The spectrum dN/dω = (2α/π)f_IR/ω integrates to N = (α/π)f_IR ln(ω₂²/ω₁²) = 0.0023228 × 13.43 × ln 100 = 0.0023228 × 13.43 × 4.605 = 0.144.
The energy is ∫ω dN = (2α/π)f_IR(ω₂ − ω₁) = 0.0046456 × 13.43 × 45 MeV = 2.81 MeV, the flat energy spectrum times the width of the band.
Checks: N is a pure number, as α, f_IR and a logarithm are. About one scatter in seven carries such a photon; the energy radiated averages 0.14% of the beam's per scatter, and even a 50 MeV photon takes only 2.5%, so leaving the electron's kinematics unchanged is consistent.

Answerf_IR = 13.43; a mean of 0.144 photons per scatter between 5 and 50 MeV, carrying 2.8 MeV between them.

Medium1.0 GeV electrons scatter through 30° from a recoil-free heavy nucleus, and a 2.0 MeV photon leaves in the scattering plane at 90° to the beam, on the side the electron was deflected towards. Build J^μ, check k⋅J = 0, compare the sum over the two physical polarisations with −J⋅J, and find dN/(dω dΩ) there against its average over directions.

Take the beam along z, the deflection towards +x and the photon along +x, so k = ω(1, 1, 0, 0) with ω = 0.0020 GeV, and E = |p| = 1.0 GeV to 1 part in 10⁷. Then p⋅k = ωE = 0.0020 GeV² and p′⋅k = ωE(1 − sin 30°) = 0.0010 GeV².
J^μ = p′^μ/(p′⋅k) − p^μ/(p⋅k): J⁰ = 1000 − 500 = 500, J¹ = 0.5/0.0010 = 500, J² = 0 and J³ = 0.8660/0.0010 − 1/0.0020 = 366.0, all in GeV⁻¹. Then k⋅J = ω(J⁰ − J¹) = 0: the timelike and longitudinal components are equal.
The physical polarisations for k along x are ŷ and ẑ: ŷ gives 0 and ẑ gives (J³)² = 1.340 × 10⁵ GeV⁻². The −g_μν sum gives −J⋅J = (J¹)² + (J³)² − (J⁰)² = 1.340 × 10⁵ GeV⁻² as well, its timelike −2.5 × 10⁵ and longitudinal +2.5 × 10⁵ cancelling.
dN/(dω dΩ) = (α/(4π²)) ω (−J⋅J) = 1.8484 × 10⁻⁴ × 0.0020 GeV × 1.340 × 10⁵ GeV⁻² = 0.0495 GeV⁻¹ sr⁻¹, or 4.95 × 10⁻⁵ MeV⁻¹ sr⁻¹: a number per unit energy per steradian, as it must be.
Here ω²(−J⋅J) = 0.536, against its average over directions 2f_IR = 25.68, so this wide-angle direction gets 2.1% of the mean density: most photons sit in cones of width 1/γ = 0.03° about p and p′. The photon here is fully polarised in the scattering plane.

AnswerJ^μ = (500, 500, 0, 366) GeV⁻¹ with k⋅J = 0; both sums give 1.34 × 10⁵ GeV⁻²; dN/(dω dΩ) = 4.95 × 10⁻⁵ MeV⁻¹ sr⁻¹, 2.1% of the direction average, polarised in the plane.

HardAn electron, a muon and a negative pion, each of momentum 500 MeV, scatter through 60° from a recoil-free heavy nucleus. Use f_IR = χ coth χ − 1 to find the mean number of photons each emits between 1 and 10 MeV, test the large −q² and dipole forms on the muon, and explain why the pion's zero spin does not enter.

All three have −q² = 4|p|² sin²30° = 0.2500 GeV², so |q| = 500 MeV and every photon up to 10 MeV is soft. Their energies are 500.0, 511.0 and 519.1 MeV, so γ = 978, 4.84 and 3.72.
cosh χ = 1 − q²/(2m²). Electron: 4.787 × 10⁵, χ = 13.772, f_IR = 12.772. Muon: −q²/m² = 22.39, cosh χ = 12.197, χ = 3.1927, coth χ = 1.00338, f_IR = 2.2034. Pion: −q²/m² = 12.83, cosh χ = 7.4169, χ = 2.6923, coth χ = 1.00922, f_IR = 1.7171.
N = (α/π)f_IR ln 100 = 0.010697 f_IR: 0.1366 for the electron, 0.0236 for the muon and 0.0184 for the pion.
For the muon, ln(−q²/m²) − 1 = 2.109 is 4.3% low, because −q²/m² = 22 is not yet large, and the dipole form −q²/(3m²) = 7.46 is 3.4 times too high, because v = 0.978 is not slow. Neither limit holds, and the closed form is needed.
The soft factor holds only charge and four-momentum, so the spinless pion radiates exactly as a spin-½ particle of 139.6 MeV would. It radiates less than the muon only because, at equal momentum, it is slower, with wider cones, 1/γ = 15° against 12°, and a smaller collinear logarithm.
The electron's f_IR exceeds the muon's by 10.57, close to ln(m_μ²/mₑ²) = 10.66: the collinear logarithm of cones only 0.06° wide, which makes the electron radiate 5.8 times as much.

AnswerN = 0.137 for the electron, 0.0236 for the muon and 0.0184 for the pion; for the muon the large −q² form is 4% low and the dipole form 3.4 times high; spin never enters the soft factor.